### Video Transcript

Determine the integral of the cos
of seven π₯ minus seven with respect to π₯.

The question is asking us to
evaluate the integral of a composite function. Itβs the integral of the
composition of a trigonometric function and a linear function. And thereβs a lot of different ways
we could think about evaluating this integral. For example, we might want to try
using our multiple angle formulas to rewrite our integrand. However, this is also the integral
of a composite function. So we should also check if thereβs
a suitable π’ substitution we can use to make this integral easier. And in this case, this is true. Weβll set π’ to be our inner
function. Thatβs the linear function seven π₯
minus seven. Then, we can differentiate π’ with
respect to π₯. Since this is a linear function, we
just get the coefficient of π₯ which is seven.

At this point, we need to be a
little bit careful. Normally, we would check to see if
dπ’ by dπ₯ appears in our integrand so we could use this to make or integral
simpler. However, on first glance, this
doesnβt seem to be the case. However, if we multiply our
integrand by seven and then multiply our entire integral by one over seven, we donβt
change the value of our integral. But now we have dπ’ by dπ₯
appearing in our integrand. And in doing this, weβve rewritten
our integral to be of the form one-seventh times the integral of π prime of π₯
times π of π of π₯ with respect to π₯. We can use the reverse chain
rule.

So letβs try and evaluate our
integral by using the substitution π’ is equal to seven π₯ minus seven. Weβve already shown dπ’ by dπ₯ is
equal to seven. And remember, dπ’ by dπ₯ is not a
fraction. However, when weβre integrating by
substitution, we can treat it a little bit like a fraction. This gives us the equivalent
statement in terms of differentials dπ’ is equal to seven dπ₯. Weβre now ready to evaluate our
integral by using substitution. First, the integral of the cos of
seven π₯ minus seven with respect to π₯ is equal to one-seventh times the integral
of seven times the cos of seven π₯ minus seven with respect to π₯.

Next, we set π’ to be our inner
function seven π₯ minus seven. We then show that dπ’ by dπ₯ is
equal to seven. And this is equivalent to saying
dπ’ is equal to seven dπ₯. In other words, in our integration
by substitution, we can replace seven dπ₯ with dπ’. So using integration by
substitution, we get one-seventh multiplied by the integral of the cos of π’ with
respect to π’. And we know how to evaluate this
integral. The integral of the cos of π with
respect to π is equal to the sin of π plus the constant of integration π. So applying this rule with π’
instead of π, we get one-seventh times the sin of π’ plus the constant of
integration weβll call π one.

Distributing one-seventh over our
parentheses, we get one-seventh times the sin of π’ plus π one over seven. And weβll simplify our answer. First, π one is a constant. So π one divided by seven is also
a constant. So weβll just replace this with a
new constant weβll call π. Next, remember, our original
integral was in terms of π₯. So we want to give our answer in
terms of π₯. Weβll do this by reusing our
substitution π’ is equal to seven π₯ minus seven. This gives us one-seventh times the
sin of seven π₯ minus seven plus π. And this is our final answer.

Therefore, by using integration by
substitution, weβve shown the integral of the cos of seven π₯ minus seven with
respect to π₯ is equal to one-seventh times the sin of seven π₯ minus seven plus our
constant of integration π.